Girl in trouble

[quiet]

Hi. One girl in high school said the following: It bothers me a lot that the relativistic kinetic energy formula has [math] v ^ 2 [/math] tucked into a root that is in the denominator. That prompted me to look for a way to rewrite that relativistic equation in a way that has more Newtonian flavor. Welcome your didactic opinions, or the type that they are. Is the next.

[math] T = m_o \ C^2 \left( \dfrac{1}{ \sqrt{ 1 - \dfrac{v^2}{C^2}} } \ -1\right) [/math]

[math] T = m_o \ \ C^2 \ \dfrac{1- \sqrt{ 1 - \dfrac{v^2}{C^2}} }{ \sqrt{ 1 - \dfrac{v^2}{C^2}} } [/math]

[math] T = m_o \ \ C^2 \ \dfrac{1- \sqrt{ 1 - \dfrac{v^2}{C^2}} }{ \sqrt{ 1 - \dfrac{v^2}{C^2}} } \ \ \dfrac{1+ \sqrt{ 1 - \dfrac{v^2}{C^2}} }{ 1+\sqrt{ 1 - \dfrac{v^2}{C^2}} } [/math]

We solve the difference of squares in the numerator.

[math] T = m_o \ \ C^2 \ \dfrac{1-1+\dfrac{v^2}{C^2}}{ \sqrt{ 1 - \dfrac{v^2}{C^2}} \left(1+\sqrt{ 1 - \dfrac{v^2}{C^2}} \right) } [/math]

We simplify and order.

[math] T = \dfrac{m_o \ v^2}{ \sqrt{ 1 - \dfrac{v^2}{C^2}} \left(1+\sqrt{ 1 - \dfrac{v^2}{C^2}} \right) } [/math]

Let's symbolize R to the denominator.

[math] R =  \sqrt{ 1 - \dfrac{v^2}{C^2}} \left(1+\sqrt{ 1 - \dfrac{v^2}{C^2}} \right) [/math]

When the speed tends to zero we have the following.

[math] \lim\limits_{\large{v \to 0}}(R)=2 [/math]

I suppose that in that way the girl will feel less rejection for relativistic kinetic energy. Is that likely or have I complicated everything in a worse way?

Edited August 22, 2018 by quite

[Markus Hanke]

8 hours ago, quite said:

It bothers me a lot that the relativistic kinetic energy formula has v2 tucked into a root that is in the denominator. 

Why would that be a problem?

 

[quiet]

The girl started asking what part of the Newtonian formula fails at great speeds. I answered the following. The equation of kinetic energy is not established by choice. It follows from the structure of the theory. Are you checking that this equation does not work for high speeds? Then the whole theory, with all its structure, does not work for great speeds. He judged that response as evasive and did not accept any further dialogue. That's why I wanted to present the relativistic formula in a way that the girl could associate with the Newtonian formula.

Today I showed her the same thing that I wrote in this note. He loved it and said that now he could understand what is the failed part of the Newtonian formula. Instead of contradicting her I preferred to ask what is the failed part. It's 2, he replied. And she was very happy, although as ignorant as before regarding the characteristics of physical theories. The important thing is that the dialogue was reopened. Now only lack she understands the structural flank.

[DimaMazin]

We can do the formula without c .

gamma=1/(1-c²/v²)^1/2

then         c²=gamma²v²/(gamma²-1)

then         kinetic energy=m₀(gamma-1)gamma²v²/(gamma²-1)

kinetic energy =m₀(gamma -1)gamma²v²/[(gamma-1)(gamma+1)

kinetic energy=m₀*gamma²v²/(gamma+1)

In newtonian physics gamma=1

therefore    kinetic energy=m₀*1²v²/(1+1)